$f(n) = -2n^{2}+6n-4(g(n))$ $g(n) = -3n$ $h(t) = -6t^{3}-6t^{2}+2(g(t))$ $ f(g(3)) = {?} $
Answer: First, let's solve for the value of the inner function, $g(3)$ . Then we'll know what to plug into the outer function. $g(3) = (-3)(3)$ $g(3) = -9$ Now we know that $g(3) = -9$ . Let's solve for $f(g(3))$ , which is $f(-9)$ $f(-9) = -2(-9)^{2}+(6)(-9)-4(g(-9))$ To solve for the value of $f$ , we need to solve for the value of $g(-9)$ $g(-9) = (-3)(-9)$ $g(-9) = 27$ That means $f(-9) = -2(-9)^{2}+(6)(-9)+(-4)(27)$ $f(-9) = -324$